This preview shows page 1. Sign up to view the full content.
Unformatted text preview: + 2nd 1. Step Input order G(s) A sudden change in a process variable can be approximated by
a step change of magnitude, M: In analyzing process dynamic and process control systems, it is
important to know how the process responds to changes in the
important to know how the process responds to changes in the
process inputs. 1storder response A number of standard types of input changes are widely used for
number of standard
of input changes are widely used for
two reasons:
1. They are representative of the types of changes that occur
in real operating processes. Chapter 5 Chapter 5 Dynamic Behavior: 1st 2. They are easy to analyze mathematically. Us ⎧0 t<0
⎨
⎩M t ≥ 0 (54) The step change occurs at an arbitrary time denoted as t = 0.
step change occurs at an arbitrary time denoted as
0.
• Special Case: If M = 1, we have a “unit step change”. We
give it the symbol,
give it the symbol, S(t).
• Example of a step change: A reactor feedstock is suddenly
switched from one supply to another, causing sudden
changes in feed concentration, flow, etc. 1 2 Example: We can approximate a drifting disturbance by a ramp input: and Q ( t ) = 8000 + 2000S ( t ) , S (t ) Q′ ( t ) = Q − Q = 2000 S ( t ) , 1storder response U R (t ) Q = 8000 kcal/hr unit step 2. Ramp Input
• Industrial processes often experience “drifting
disturbances”, that is, relatively slow changes up or down
for some period of time.
• The rate of change is approximately constant. The deviation
variable is UR(t).
3 Chapter 5 Chapter 5 The heat input to the stirredtank heating system in Chapter 2 is
heat input to the stirred
heating system in Chapter is
suddenly changed from 8000 to 10,000 kcal/hr by changing the
electrical signal to the heater. Thus, ⎧0 t < 0
⎨
⎩a it t ≥ 0 (57) Examples of ramp changes:
of ramp changes:
1. Ramp a setpoint to a new value. (Why not make a step
change?)
change?)
2. Feed composition, heat exchanger fouling, catalyst
activity, ambient temperature.
3. Rectangular Pulse Input It represents a brief, sudden change in a process variable:
represents brief sudden change in process variable:
4 U RP ( t ) ⎧0
⎪
⎨h
⎪0
⎩ for t < 0
for 0 ≤ t < tw (59) for t ≥ tw Chapter 5 Chapter 5 XRP
h 0 tw Time, t Fig. 5.2 Three important
examples of deterministic
inputs. Examples: 1. Reactor feed is shut off for one hour.
2. The fuel gas supply to a furnace is briefly interrupted.
5 6 5. Impulse Input 4. Sinusoidal Input (later = Chap. 14, Sinusoidal Ouput)
Processes are also subject to periodic, or cyclic, disturbances.
They can be approximated by a sinusoidal disturbance: ⎧0 for t < 0
(514)
⎨
⎩ A sin (ωt ) for t ≥ 0
2ndorder response
may have increased
amplitude at some frequencies! where: Here, U I ( t ) = δ ( t ) .
It represents a short, transient disturbance. Examples of approximate impulses:
approximate impulses: Chapter 5 Chapter 5 U sin ( t ) •
• A = amplitude, ω = angular frequency, 1/time
amplitude,
angular frequency, 1/time
= (radians)/time not 1. Electrical noise spike in a thermocouple reading.
2. Injection of a tracer dye.
• Useful for analysis since the time response to an impulse
input is the inverse Laplace transform of the TF Thus
input is the inverse Laplace transform of the TF. Thus,
u (t ) U (s) (cycles)/time
Examples of “cycles”:
1. 24 hour variations in cooling water temperature.
2. 60Hz electrical noise (in the USA) Here,
7 G (s) y (t ) Y (s) Y ( s ) = G ( s )U ( s ) (1)
8 FirstOrder System
Th
The corresponding time domain express is:
0 Chapter 5 where: g ( t ) L−1 ⎡G ( s ) ⎤
⎣
⎦ Y (s) (2) Chapter 5 t y ( t ) = ∫ g ( t − τ ) u ( τ ) dτ The standard form for a firstorder TF is:
fi
TF (3) Suppose u ( t ) = δ ( t ) . Then it can be shown that:
y (t ) = g (t ) (4) Consequently, g(t) is called the “impulse response function”. where: U (s)
K
τ = K
τs + 1 (516) steadystate gain
time constant Consider the response of this system to a step of magnitude, M:
U ( t ) = M for t ≥ 0 ⇒ U (s) = M
s Substitute into (516) and rearrange,
Y (s) = KM
s ( τs + 1) (517) 9 10 Integrating Process Take L1 (cf. Table 3.1) = memorize it ! , ( y ( t ) = KM 1 − e−t / τ steadystate value of y(t). From (518), y∞ = KM . 1.0 y
y∞ 0.5 0 0 Not all processes have a steadystate gain. For example, an
“integrating process” or “integrator” has the transfer function: (518) 1 2 3 t
τ 4 5 t
0
τ
2τ
3τ
4τ
5τ Chapter 5 Chapter 5 Let y∞ )
y
y∞
___ 0
0.632
0.865
0.950
0.982
0.993* nearly complete Note: Larger τ means a slower response. 11 Y (s) U (s) = K
s ( K = constant ) Consider a step change of magnitude M. Then U(s) = M/s and,
1
KM L
Y ( s ) = 2 ⇒ y ( t ) = ( KM ) t
s
Thus, y(t) is unbounded and a new steadystate value does not
exist, i.e. there is no Final Value ! 12 Common Physical Example: SecondOrder Systems Consider a liquid storage tank with a pump on the exit line:
li
li • Standard form:
Y (s)  Chapter 5 Chapter 5 U (s) Assume:
1. Constant crosssectional area, A.   2. q ≠ f ( h ) this is not the check quiz story problem !
dh
Mass balance: A = qi − q ⇒ 0 = qi − q s.s. equation
dt
Eq. (1) – Eq. (2), take L, assume steady state initially,
1
⎡Qi′ ( s ) − Q′ ( s ) ⎤
H ′( s) =
⎦
H ′( s ) 1
As ⎣
=
Qi′ ( s ) As
For Q′ ( s ) = 0 (constant q), = K
22 τ s + 2ζτs + 1 (540) which has three model parameters:
K steadystate gain τ "time constant" >0 [=] time
constant" >0
time
ζ damping coefficient (dimensionless)
1⎞
⎛
• Equivalent form: ⎜ ωn natural frequency = ⎟
τ⎠
⎝
Y (s) U (s) = 2
K ωn
2
s 2 + 2ζ ω n s + ω n 13 14 • The type of behavior that occurs depends on the numerical
value of damping coefficient, ζ :
It is convenient to consider three types of stable behavior:
Type of Response Roots of Charact.
Polynomial ζ >1 Overdamped Negative real and ≠ ζ =1 Critically damped Negative real and = Underdamped Complex conjugates
negative real parts 0 ≤ ζ <1 Chapter 5 Chapter 5 Damping
Coefficient Note: this is for
this is for
dimensionless time,
“Speed” of approach
to final value depends on the
final value depends on the
time constant (tau) – and only
“somewhat” on zeta. • Note: The characteristic polynomial is the denominator of the
The characteristic polynomial the denominator of the
transfer function:
τ 2 s 2 + 2ζτs + 1 • What about ζ < 0 ? It results in an unstable system
15 16 1. Responses exhibiting oscillation and overshoot (y/KM > 1) are
obtained only for values of ζ less than one. Again  dimensionless time,
of approach
“Speed” of approach
to final value depends on the
time constant (tau) – and only
“somewhat” on zeta. Chapter 5 Chapter 5 Several general remarks can be made concerning the
responses show in previous two Figs. 5.8 and 5.9: 2. Large values of ζ yield a sluggish (slow) response for fixed τ .
3. The fastest response without overshoot is obtained for the
fastest response without overshoot obtained for the
critically damped case for fixedτ . ( ζ = 1) if the two time constants are identical. 17 18 1. Rise Time: tr is the time the process output takes to first reach
the new steadystate value. Chapter 5 Chapter 5 2. Time to First Peak: tp is the time required for the output to
reach its first maximum value.
3. Settling Time: ts is defined as the time required for the
process output to reach and remain inside a band whose width
is equal to ±5% of the total change in y. The term 95%
Th
95%
response time sometimes is used to refer to this case. Also,
values of ±1% sometimes are used.
4. Overshoot: OS = a/b (% overshoot is 100a/b).
5. Decay Ratio: DR = c/a (where c is the height of the second
Ratio DR
the height of the second
peak).
6. Period of Oscillation: P is the time between two successive
peaks or two successive valleys of the response.
19 20 Second Order Step Change
a. Block Notation: overshoot ζ >1 b. time of first maximum
⎛ πτ
tp = ⎜
⎜ 1− ζ 2
⎝ c. ⎞
⎟
⎟
⎠ decay ratio (successive maxima – not min.)
ratio (successive maxima
min
⎛ −2πζ
c
= exp ⎜
⎜ 1− ζ 2
a
⎝ ⎞ a2
⎟= 2
⎟b
⎠ d. period of oscillation
⎛ 2πτ
P=⎜
⎜ 1− ζ 2
⎝ ⎞
⎟
⎟
⎠ overdamped Composed of two first order subsystems (G1 and G2) Chapter 5 Chapter 5 ⎛ −πζ ⎞
a
= exp ⎜
⎟ depends only on ζ , the damping coefficient
2⎟
⎜
b
⎝ 1− ζ ⎠ G ( s )= K General 2ndorder K ζ= τ1 + τ 2
2 τ1τ 2 K
D( s)
Roots of D( s ) = 0 are poles of G ( s ). G (s )= 2 "poles" are at {s1 , s2 } =
21 τ = τ 1τ 2 τ 1τ 2 s 2 +(τ 1 + τ 2 )s +1 τ 2 s 2 + 2ζτ s + 1 −ζ ± ζ 2 − 1 τ = ζ <1
ζ =1 underdamped
critically damped 22 ...
View
Full
Document
This note was uploaded on 03/01/2012 for the course CHE 361 taught by Professor Staff during the Winter '08 term at Oregon State.
 Winter '08
 Staff

Click to edit the document details